6 Home exercise sheet 6.1 Hamilton Jacobi and the central force problem

6
6.1
Home exercise sheet
Hamilton Jacobi and the central force problem
Exercise 4.1: Hamilton Jacobi
A particle with mass m is moving under the given potential
1
V (r, φ, z) = log z 2 (f (φ))− r2
(1)
Write the Hamilton-Jacobi equation and find the constants of motion.
Solution:
The Hamiltonian of the problem is
H=
p2φ
1
p2r
p2
+
+ z + log z 2 (f (φ))− r2
2m 2m 2m
So, if we open the log, the H-J equation will be
!
2
2
1
∂s
1
∂s
1
+ 2 log z +
+ 2
2m ∂z
2m ∂r
r
1
2m
∂s
∂φ
(2)
!!
2
− log (f (φ))
=E
(3)
Notice that the first bracket is only dependent on z and the second bracket is only dependent
on φ and r. We again use separation of variables giving us
2
∂s
1
+ 2 log z = Ez
(4)
2m ∂z
!
2
2
∂s
1
1
∂s
1
+ 2
− log (f (φ))
= Erφ
(5)
2m ∂r
r
2m ∂φ
Let us look at Erφ . The inner brackets are only φ dependent so choose
w(r, φ) = wr (r) + wφ (φ)
(6)
This will give us
1
2m
∂wr
∂r
2
1
+ 2
r
1
2m
∂wφ
∂φ
1
2
!
− log (f (φ))
= Erφ
(7)
Since the inner brackets only depend on φ we can write
2
∂wφ
1
− log (f (φ))
⇒
const = αφ =
2m ∂φ
We can now replace
dwi
dxi
1
2m
∂wr
∂r
2
+
αφ
= Erφ
r2
(8)
with pi and receive our constants of motion
Ez
αφ
Erφ
p2z
+ 2 log z
=
2m
p2φ
=
− log (f (φ))
2m
p2r
αφ
=
+ 2
2m
r
(9)
(10)
(11)
Exercise 4.2: Laplace-Runge-Lenz vector
For a motions in central field. Prove that the following vector is conserved.
A = p × l − µkr̂
(12)
What is its direction? What is it’s magnitude?
Solution:
To show that a vector is a constant we just need to derive it, in our case we will derive it in
time.
rṙ − r ṙ
dA
= ṗ × l + p × l̇ − µk
(13)
dt
r2
kr
ṙ
ṙr
= − 3 × (µr × ṙ) − µk + µk 2
(14)
r
r
r
r (r · ṙ)
ṙ (r · r)
ṙ
ṙr
= −µk
+ µk
− µk + µk 2 = 0
(15)
3
3
r
r
r
r
So A is a conserved vector which clearly lies in the plane of the motion. A points toward
periapsis, i.e. toward the point of closest approach to the force center.
The magnitude of the vector is as follows:
A2 = (p × l)2 − 2µkr̂ · p × l + µ2 k 2
k
= p2 l2 − 2µl2 + µ2 k 2
2 r
p
k µk 2
µk 2
2
2
= 2µl
− + 2 = 2µl E + 2
2µ r
2l
2l
2
(16)
(17)
(18)
We can now plug in the relation between and E which you found in class and get.
A2 = (µk)2
(19)
Exercise 4.3: Mass on a conic surface
A mass m slides without friction on the inside of a conical surface with an opening angle α.
1. Find the Lagrangian for the mass using cylindrical coordinates r, θ, z. Use the constraint to eliminate the z dependence.
2. Is there a conserved quantity in the problem? Use them to reduce the problem to a
single dynamical variable.
3. Show that the solution can be described by motion of a particle with an effective mass
in a one-dimensional effective potential. What are the effective mass and the effective
potential?
4. Use the effective potential you have obtained to calculate the radius of a circular orbit.
5. What is the period of the circular orbit?
6. Calculate the frequency of small oscillations around the circular orbit.
Solution:
3
1. The Lagrangian for the mass is:
1 L = m ṙ2 + r2 θ̇2 + ż 2 − mgz
2
On the conical surface z = r cot α. Using this relation we have:
1 L = m ṙ2 1 + cot2 α + r2 θ̇2 − mgrcotα
2
(20)
(21)
2. The angular momentum is a conserved quantity since θ is a cyclic coordinate. The
energy is also a conserved quantity since the Lagrangian is time independent. Using
this we can write the energy of the problem with a single dynamical variable,
1
L2
E = m 1 + cot2 α ṙ2 +
+ mgr cot α
(22)
2
2mr2
3. This corresponds to a motion of a single particle with mass m(1 + cot2 α) in a one
dimensional effective potential:
Vef f =
L2
+ mgr cot α
2mr2
(23)
4. The particle performs a circular orbit when the energy is equal to the minimum of the
effective potential (corresponding to an orbit with a constant radius). The radius of the
circular orbit is therefore:
dVef f (rmin )
L2
=
+ mg cot α = 0
3
dr
mrmin
13
L2
rmin =
m2 g cot α
(24)
(25)
5. The period of the orbit can be found by inverting the relation for θ(t) as implied from
the conservation of angular momentum, and integrating over a period:
Z T
Z 2π
2
mrmin
mr2
T =
dT =
dθ = 2π min
(26)
L
L
0
0
6. The frequency of small oscillations around the circular orbit is given by expanding the
effective potential to second order:
s
s
s
d2 Vef f (rmin )
k
3L2
2
dr
=
=
(27)
Ω=
4
mef f
mef f
m2 rmin
(1 + cot2 α)
4